15.6 Triple integrals

• The following topics will be covered in this lecture:
  • Triple integrals over rectangular boxes
  • Fubini’s theorem for triple integrals
  • Triple integrals over non-rectangular solids

• Suppose you want to know the mass density of an object with a density function of three variables \( f(x,y,z) \) \[ f : \mathbb{R}^3 \rightarrow \mathbb{R} \]
• The domain of the function \( E \) is a subset of \( \mathbb{R}^3 \) \[ E \subseteq \mathbb{R}^3 \]
• Suppose that \( E \) is a bounded region in three dimensional space (a solid).

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• Let’s first deal with the simplest case where \( f \) is defined on a rectangular box \( B \)
• The first step is to divide \( B \) into sub-boxes.
• Each sub-box \( B_{ijk} \) has volume \[ \Delta V_{ijk} = \Delta x_i \Delta y_j \Delta z_k \]
• By analogy with the definition of a double integral, we define the triple integral as the limit of the triple Riemann sums

  \[ \iiint_\limits{B} f(x,y,z)dV = \lim_{\Delta x_i \rightarrow 0}\lim_{\Delta y_j \rightarrow 0}\lim_{\Delta z_k \rightarrow 0} \sum_{i=1}^l\sum_{j=1}^m \sum_{k=1}^n f\left( x^*_{ijk}, y^*_{ijk}, z^*_{ijk}\right)\Delta V_{ijk} \]

  if the limit exists

• Evaluate the triple integral \( \iiint \limits_B \left(x+yz^2 \right)dV \) where \( B \) is the rectangular box given by \[ B= \{(x,y,z)|\;-1\leq x \leq 5,\: 0\leq y \leq 4,\: 0\leq z \leq 1 \} \]

• Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals \[ I=\iiint \limits_B \left(x+yz^2 \right)dV = \int_{{\color{purple} {z=0}}}^{{\color{purple} {z=1}}} \left\{ \int_{{\color{blue} {y=0}}}^{{\color{blue} {y=4}}} \left [\int_{{\color{red} {x=-1}}}^{{\color{red} {x=5}}} \left(x+yz^2 \right) {\color{red} {dx}}\right] {\color{blue} {dy}} \right\} {\color{purple} {dz}} \]
• Start integrating from inside to outside
• Integrate with respect to \( x \) while holding \( y \) and \( z \) constant.
  \( I= \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left\{ \int_{{\color{blue} {0}}}^{{\color{blue} {4}}} \left(\frac{x^2}{2}+yz^2x \right)\bigg\vert_{{\color{red} {x=-1}}}^{{\color{red} {x=5}}} {\color{blue} {dy}} \right\} {\color{purple} {dz}} \)
• \( I= \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left\{ \int_{{\color{blue} {0}}}^{{\color{blue} {4}}} \left(\frac{25}{2}+5yz^2-\left(\frac{1}{2}-yz^2\right) \right) {\color{blue} {dy}} \right\} {\color{purple} {dz}} \)
• \( I= \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left\{ \int_{{\color{blue} {0}}}^{{\color{blue} {4}}} \left(12+6yz^2 \right) {\color{blue} {dy}} \right\} {\color{purple} {dz}} \)

• Finish evaluating the remaining double integral
• \[ = \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left[ \left(12y+3y^2z^2 \right) \bigg\vert_{{\color{blue} {y=0}}}^{{\color{blue} {y=4}}} \right] {\color{purple} {dz}} \]
• \[ = \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left[ \left(12(4)+3(4)^2z^2 \right) \right] {\color{purple} {dz}} \]
• \[ = 48 \int_{{\color{purple} {0}}}^{{\color{purple} {1}}} \left(1+z^2 \right) {\color{purple} {dz}} \]
• \[ \begin{align} &= 48\left(z+\frac{z^3}{3} \right)\bigg\vert_{{\color{purple} {z=0}}}^{{\color{purple} {z=1}}}\\ &=48\left(1+\frac{1}{3} \right)\\ &=48\left(\frac{4}{3} \right) =64 \end{align} \]
• Therefore \[ \iiint_B \left(x+yz^2 \right)dV = 64 \]

• If \( f \) is continuous on the rectangular box \[ B=[a,b]\times[c,d]\times[r,s] \] then \[ \iiint \limits_B f(x,y,z)dV= \int_r^s \int_c^d \int_a^b f(x,y,z) dx dy dz \]

• This integral is also equal to any of the other five possible orderings for the iterated triple integral

• dxdzdy
  dydxdz
  dydzdx
  dzdxdy
  dzdydx
  

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• Now we define the triple integral over a general bounded region \( E \) in \( \mathbb{R}^3 \) (a solid)
• Enclose \( E \) in a box \( B \). Then we define a function \( F \) so that it agrees with \( f \) on \( E \) but is \( 0 \) for points in \( B \) that are outside \( E \)
• \[ \iiint \limits_E f(x,y,z) dV = \iiint \limits_B F(x,y,z)dV \]
• Solid of type I : if it lies between the graphs of two continuous functions of \( x \) and \( y \), that is \[ E= \{ (x,y,z)| (x,y)\in D,\; u_1(x,y) \leq z \leq u_2(x,y) \} \]
• where \( D \) is the projection of \( E \) onto the \( xy \)-plane as shown in Figure.
• As \( E \) is a type 1 solid, then \[ \iiint \limits_E f(x,y,z)dV = \iint_\limits D \left[\int_{{\color{purple} {u_1(x,y)}}}^{{\color{purple} {u_2(x,y)}}} f(x,y,z) {\color{purple} {dz}} \right]dA \]

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• In particular, if the projection \( D \) of \( E \) onto the \( xy \)-plane is a type I plane region
  
  \[ E=\{(x,y,z)| a\leq x \leq b, g_1(x)\leq y \leq g_2(x), \;u_1(x,y)\leq z \leq u_2(x,y) \} \]
• 
  Then the triple integral becomes
  \[ \iiint_E f(x,y,z)dV = \int _{\color{blue} {a}}^{\color{blue} {b}} \int_{{\color{green} {g_1(x)}}}^{{\color{green} {g_2(x)}}} \left[\int_{{\color{purple} {u_1(x,y)}}}^{{\color{purple} {u_2(x,y)}}} f(x,y,z) {\color{purple} {dz}} \right]{\color{green} {dy}}\; {\color{blue} {dx}} \]

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• If the projection \( D \) of \( E \) onto the \( xy \)-plane is a type II plane region
  
  \[ E=\{(x,y,z)| c\leq y \leq d, h_1(y)\leq x \leq h_2(y), \;u_1(x,y)\leq z \leq u_2(x,y) \} \]
• 
  Then the triple integral becomes
  \[ \iiint \limits_E f(x,y,z)dV = \int_{\color{green} c}^{\color{green} d} \int_{{\color{blue} {h_1(y)}}}^{{\color{blue} {h_2(y)}}} \left[\int_ {{\color{purple} {u_1(x,y)}}}^{{\color{purple} {u_2(x,y)}}} f(x,y,z) {\color{purple} {dz}} \right]{\color{blue} {dx}}\; {\color{green} {dy}} \]

• Evaluate \( \iiint \limits_{E} z dV \), where \( E \) is the solid tetrahedron bounded by the four planes \( x= 0, y= 0, z=0 \), and \( x + y + z =1 \) .

• Sketch two diagrams: one of the solid region \( E \) and one of its projection \( D \) on the \( xy \) -plane
• Lower boundary of the tetrahedron is the plane \[ z = 0 \] Upper boundary is the plane \[ x +y + z = 1 \] \[ \Leftrightarrow z = 1 - x - y \]
• So we use \[ u_1(x,y)=0 \] and \[ u_2(x,y) = 1 - x - y \]
• The planes \( x +y + z=1 \) and \( z=0 \) intersect in the line \[ x + y = 1 \] \[ \Leftrightarrow y= 1 - x \] in the \( xy \)-plane.

• Set up triple integrals over a type I solid \[ \iiint \limits_E z dV = \iint \limits_{D} \left[\int^{{\color{purple} {z=1-x-y}}}_{{\color{purple} {z=0}}} z\; {\color{purple} {dz}} \right]dA \]
• The description of E as a type 1 solid enables us to evaluate the integral as follows \[ = \int_{0}^1 \int_{0}^{1-x}\left[\int^{{\color{purple} {z=1-x-y}}}_{{\color{purple} {z=0}}} z\; {\color{purple} {dz}} \right]dy\;dx \]
• Integrate with respect to \( z \) \[ = \int_{0}^1 \int_{0}^{1-x}\left[ \frac{z^2}{2}\right]\bigg\vert^{{\color{purple} {z=1-x-y}}}_{{\color{purple} {z=0}}} \;dy\;dx \]
• Double integral over \( D \) \[ =\frac{1}{2}\int_{0}^1 \int_{0}^{1-x}\left(1-x-y\right)^2dy\;dx \]
• U-subs \( U=1-x-y \) \[ =\frac{1}{2}\int_{0}^1 \left[-\frac{(1-x-y)^3}{3}\right] \bigg\vert_{y=0}^{y=1-x}\;dx \]
• \[ =\frac{1}{6}\int_{0}^1 \left(1-x\right)^3\;dx=\frac{1}{6} \left[ -\frac{(1-x)^4}{4}\right]\bigg\vert_0^1=\frac{1}{24} \]

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• If the projection \( D \) of \( E \) onto the \( yz \)-plane is a plane region \( D \)
  
  
• A solid region \( E \) is of type II if it is of the form \[ E=\{(x,y,z)| (y,z)\in D, \;u_1(y,z)\leq x \leq u_2(y,z) \} \]
• 
  Then the triple integral becomes
  \[ \iiint \limits_E f(x,y,z)dV = \iint \limits_D \left[\int_ {{\color{blue} {u_1(y,z)}}}^{{\color{blue} {u_2(y,z)}}} f(x,y,z) {\color{blue} {dx}} \right] \; dA \]

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

• If the projection \( D \) of \( E \) onto the \( xz \)-plane is a plane region \( D \)
  
  
• A solid region \( E \) is of type III if it is of the form \[ E=\{(x,y,z)| (x,z)\in D, \;u_1(x,z)\leq y \leq u_2(x,z) \} \]
• 
  Then the triple integral becomes
  \[ \iiint \limits_E f(x,y,z)dV = \iint \limits_D \left[\int_ {{\color{green} {u_1(x,z)}}}^{{\color{green} {u_2(x,z)}}} f(x,y,z) {\color{green} {dy}} \right] \; dA \]

• Evaluate \( \iiint \limits_{E} \sqrt{x^2+z^2} dV \), where \( E \) is the region bounded bby the paraboloid \( y=x^2+z^2 \) and the plane \( y=4 \)

• Which type of solid?
• The projection of \( E \) onto the xz -plane is the disk \( x^2 + z^2\leq 4 \)
• Regard \( E \) as a type III solid \[ \iint\limits_{Disk} \left[ \int_{y=x^2+z^2}^{y=4} \sqrt{x^2+z^2} \;dy \right]dA \]
• \[ E=\{(x,y,z)|\;(x,z)\in Disk,\;x^2+z^2 \leq y\leq 4\} \]

• Integrate with respect to \( y \) \[ =\iint\limits_{Disk} \left[y \sqrt{x^2+z^2} \right] \bigg\vert_{y=x^2+z^2}^{y=4} dA \]
• Evaluate\[ =\iint\limits_{Disk} \left[4 \sqrt{x^2+z^2} - (x^2+z^2) \sqrt{x^2+z^2}\right]dA \]
• Simplify\[ =\iint\limits_{Disk} \left[(4-x^2-z^2) \sqrt{x^2+z^2} \right]dA \]

• Easier to use polar coordinates on the \( xz \)-plane!
  \[ x=r\cos\theta\;\;z=r\sin\theta \]
• \[ =\int_0^{2\pi}\int_0^2 \left[(4-r^2)r \right] \;r\;dr\;d\theta \]
• Integrate over a polar rectangle! \[ =\int_0^2 \left(4r^2-r^4 \right) \;dr\int_0^{2\pi}d\theta=\dots=\frac{128\pi}{15} \]

• We can use Fubini’s theorem to compute a triple integral over a rectangular box.
• Recognize when a function of three variables is integrable over a closed and bounded solid
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• Questions